Adding an Initial Delay
In an ordinary renewal process, the first renewal occurs after the first interarrival time. Some models instead have a separate waiting time before the renewal pattern begins: a component may already be in use at the start of observation, or a system may begin operating only after a setup period. After that initial delay, the successive gaps follow the usual renewal law. The expected number of renewals is then governed by a measure that combines the delay law with the ordinary renewal measure.
Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Let \(X_1,X_2,\ldots\) be independent and identically distributed, strictly positive random variables with common law \(F\), and define \(S_0=0\) and \(S_n=X_1+\cdots+X_n\) for \(n\ge1\). Let \(D\) be a finite, nonnegative random variable independent of the entire sequence \((X_n)_{n\ge1}\). The delayed renewal epochs are \(R_n=D+S_n\), \(n\ge0\). We count an epoch at time zero if one occurs; in particular, if \(D=0\), then \(R_0=0\) is counted.
Both measures are locally finite: each bounded time interval has finite measure. This property is not a consequence merely of there being finitely many renewals almost surely; an almost surely finite count can still have infinite expectation. Here a direct expectation bound gives the needed stronger conclusion.
Proof. Since \(X_1>0\) almost surely, at least one of the events \(\{X_1\ge 1/m\}\), \(m\ge1\), has positive probability: their union is \(\{X_1>0\}\). Choose \(\varepsilon>0\) with \(p=\mathbb{P}(X_1\ge\varepsilon)>0\). Fix \(t\ge0\), let \(q=\lfloor t/\varepsilon\rfloor+1\), and let \(T_q\) be the index of the \(q\)-th increment at least \(\varepsilon\). If \(S_n\le t\), at most \(\lfloor t/\varepsilon\rfloor=q-1\) of \(X_1,\ldots,X_n\) can be at least \(\varepsilon\), since \(q\varepsilon>t\). Therefore \(n<T_q\). The count \(\sum_{n\ge1}\mathbf{1}_{\{S_n\le t\}}\) is at most \(T_q-1\). The indicators \(\mathbf{1}_{\{X_k\ge\varepsilon\}}\) are independent Bernoulli variables with success probability \(p\); the waiting time for each successive success has mean \(\sum_{m\ge0}(1-p)^m=1/p\). Consequently \(\mathbb{E}T_q=q/p<\infty\), and $$ U([0,t])=1+\mathbb{E}\sum_{n\ge1}\mathbf{1}_{\{S_n\le t\}} \le 1+\mathbb{E}T_q =1+\frac{q}{p}<\infty. $$ For every outcome and every \(n\), \(D+S_n\le t\) implies \(S_n\le t\), because \(D\ge0\). Hence \(N_D(t)\le\sum_{n\ge0}\mathbf{1}_{\{S_n\le t\}}\), and taking expectations gives \(U_D([0,t])\le U([0,t])<\infty\). Every compact subset of \([0,\infty)\) is contained in some \([0,t]\), proving local finiteness. \(\square\)
The Convolution Formula
Let \(F_D\) denote the probability law of \(D\). For measures on \([0,\infty)\), their convolution is defined by $$ (F_D*U)(B)=\int_{[0,\infty)}\int_{[0,\infty)} \mathbf{1}_{\{d+s\in B\}}\,U(ds)\,F_D(dd). $$ It combines all ways to make an epoch: first choose the delay \(d\), then add an ordinary renewal epoch \(s\).
Proof. Fix a Borel set \(B\subseteq[0,\infty)\). Independence of \(D\) and \(S_n\) gives, for each \(n\ge0\), $$ \mathbb{P}(D+S_n\in B) =\int_{[0,\infty)}\int_{[0,\infty)} \mathbf{1}_{\{d+s\in B\}}\,\mathbb{P}(S_n\in ds)\,F_D(dd). $$ Sum over \(n\). Every integrand is nonnegative, so Tonelli's theorem permits interchanging the sum and the integrals. The inner sum of the measures \(\mathbb{P}(S_n\in ds)\) is \(U(ds)\), by the definition of \(U\). Thus $$ U_D(B)=\int_{[0,\infty)}\int_{[0,\infty)} \mathbf{1}_{\{d+s\in B\}}\,U(ds)\,F_D(dd)=(F_D*U)(B). $$ For \(B=[0,t]\), the condition \(d+s\le t\) is impossible for \(d>t\); for \(0\le d\le t\), the allowed ordinary epochs satisfy \(s\le t-d\). Integrating their \(U\)-measure over the law of \(D\) gives the stated cumulative formula. \(\square\)
The atom at \(S_0=0\) matters. The term \(n=0\) in the delayed measure contributes the law \(F_D\), representing the first delayed epoch itself. Each later term contributes the delay law convolved with the corresponding ordinary renewal-epoch law. The formula does not say to add \(F_D\) to \(U\): the delay shifts every ordinary epoch, including the epoch at zero.
Worked Example: A Fixed Delay and Fixed Interarrivals
Suppose \(D=\tfrac12\) with probability one and \(X_n=2\) with probability one for every \(n\). These assumptions give \(S_n=2n\), so the ordinary renewal measure is \(U=\delta_0+\delta_2+\delta_4+\cdots\), where \(\delta_a\) denotes unit mass at \(a\). The delay law is \(F_D=\delta_{1/2}\).
Convolution shifts each ordinary epoch by \(\tfrac12\): $$ U_D=F_D*U=\delta_{1/2}+\delta_{5/2}+\delta_{9/2}+\cdots. $$ For \(t=\tfrac92\), the epochs satisfying \(R_n\le t\) are \(\tfrac12,\tfrac52,\tfrac92\). The inequality includes the endpoint, so \(N_D(\tfrac92)=3\) with probability one and \(U_D([0,\tfrac92])=3\).
Check: The cumulative formula gives \(U([0,\tfrac92-\tfrac12])=U([0,4])=3\), since the ordinary epochs in \([0,4]\) are \(0,2,4\). This agrees with the direct delayed count.
Worked Example: A Two-Point Delay and Two-Point Interarrivals
Let \(X_n=1\) or \(2\), each with probability \(\tfrac12\), independently across \(n\). Let \(D=0\) with probability \(\tfrac13\) and \(D=1\) with probability \(\tfrac23\), independently of all interarrivals. To compute \(U_D([0,3])\), first find the ordinary renewal masses \(u_k=U(\{k\})\) for \(k=0,1,2,3\). No epochs outside the nonnegative integers are possible.
At \(0\), only \(S_0\) contributes, so \(u_0=1\). At \(1\), only \(S_1=1\) contributes, giving \(u_1=\tfrac12\). At \(2\), \(S_1=2\) has probability \(\tfrac12\), and \(S_2=2\) when both gaps equal \(1\), with probability \(\tfrac14\). Thus \(u_2=\tfrac12+\tfrac14=\tfrac34\). At \(3\), \(S_2=3\) through the ordered pairs \((1,2)\) and \((2,1)\), total probability \(\tfrac14+\tfrac14=\tfrac12\); also \(S_3=3\) when all three gaps equal \(1\), with probability \(\tfrac18\). Hence \(u_3=\tfrac12+\tfrac18=\tfrac58\).
For integer \(k\), the convolution formula gives $$ U_D(\{k\})=\tfrac13u_k+\tfrac23u_{k-1}, $$ where \(u_{-1}=0\), since a delay of \(1\) requires an ordinary epoch at \(k-1\). Therefore the delayed masses at \(0,1,2,3\), respectively, are $$ \tfrac13,\qquad \tfrac13\cdot\tfrac12+\tfrac23\cdot1=\tfrac56,\qquad \tfrac13\cdot\tfrac34+\tfrac23\cdot\tfrac12=\tfrac7{12},\qquad \tfrac13\cdot\tfrac58+\tfrac23\cdot\tfrac34=\tfrac{17}{24}. $$ Adding them gives \(U_D([0,3])=\tfrac13+\tfrac56+\tfrac7{12}+\tfrac{17}{24}=\tfrac{59}{24}\).
Check by conditioning on \(D\): If \(D=0\), the ordinary mass through \(3\) is \(1+\tfrac12+\tfrac34+\tfrac58=\tfrac{23}{8}\). If \(D=1\), only ordinary epochs through \(2\) fit, giving \(1+\tfrac12+\tfrac34=\tfrac94\). The weighted total is \(\tfrac13\cdot\tfrac{23}{8}+\tfrac23\cdot\tfrac94=\tfrac{23}{24}+\tfrac{36}{24}=\tfrac{59}{24}\), as before.
Worked Example: A Continuous Initial Delay
Let \(D\) be uniform on \([0,1]\), and let every interarrival equal \(2\) with probability one, independently of \(D\). Then \(U=\sum_{n\ge0}\delta_{2n}\). Convolving with the uniform law shifts a copy of that law to each interval \([2n,2n+1]\); each copy has total mass one.
For \(t=\tfrac52\), the \(n=0\) delayed epoch is \(D\), which lies in \([0,\tfrac52]\) with probability \(1\). The \(n=1\) epoch is \(D+2\); it is at most \(\tfrac52\) exactly when \(D\le\tfrac12\), an event of probability \(\tfrac12\). For \(n\ge2\), \(D+2n\ge4>\tfrac52\), so those epochs contribute zero. Thus \(U_D([0,\tfrac52])=1+\tfrac12=\tfrac32\).
Check: In the cumulative convolution, \(U([0,\tfrac52-d])\) equals \(2\) for \(0\le d\le\tfrac12\), because the ordinary epochs \(0\) and \(2\) are included. It equals \(1\) for \(\tfrac12<d\le1\), because only epoch \(0\) is included. Integrating over the uniform delay gives \(\tfrac12\cdot2+\tfrac12\cdot1=\tfrac32\). The single endpoint \(d=\tfrac12\) has probability zero, but the inclusion at that endpoint is consistent with the condition \(\le\).
Interpretation and a Common Error
The convolution representation separates two sources of timing variation. The law \(F_D\) describes when the renewal pattern begins; \(U\) describes the subsequent pattern relative to its own start. Because the delay is independent, each ordinary renewal epoch is shifted by the same random \(D\), and averaging those shifted epoch laws produces \(F_D*U\). If independence fails, the joint law of \(D\) and \(S_n\) is needed; the marginal laws alone do not determine the delayed renewal measure.
Common mistake: adding the measures instead of convolving them. Take deterministic interarrivals \(X_n=2\) and deterministic delay \(D=1\). The correct delayed epochs are \(1,3,5,\ldots\), so \(U_D=\delta_1+\delta_3+\delta_5+\cdots\), and \(U_D([0,2])=1\). The incorrect expression \(U+F_D\) would put mass at \(0,1,2,\ldots\); it gives \(U([0,2])+F_D([0,2])=2+1=3\), not \(1\). It fails because it leaves the later ordinary epochs unshifted and adds the delay as a separate epoch rather than shifting the whole renewal pattern.
Another convention to check is whether the count includes its epoch at time zero. Here \(U\) includes \(S_0=0\), while \(U_D\) includes \(R_0=D\). Thus \(U_D([0,t])\) counts a first delayed epoch at zero when \(D=0\). If a problem instead defines a count that excludes time zero, the corresponding atom must be removed; one should not silently switch conventions while applying the convolution formula.
Check Your Understanding
Use the definitions and the endpoint convention in this tutorial. For each calculation, identify whether the epoch at the boundary is included.
- If \(D=0\) with probability one, what is the relation between \(U_D\) and \(U\)? Explain the role of \(S_0\).
- Suppose \(D=1\) and every interarrival equals \(3\). List the delayed epochs in \([0,7]\), and compute \(U_D([0,7])\).
- For the two-point interarrival law in the second worked example, compute \(U_D(\{4\})\) using \(U_D(\{k\})=\tfrac13u_k+\tfrac23u_{k-1}\). Which ordinary renewal masses are needed?
- Why does almost-sure finiteness of a renewal count not, by itself, establish that its renewal measure is finite on bounded intervals?
- If the initial delay is dependent on the interarrivals, which step in the convolution proof needs to be changed?